How many number of five digits can be formed without repetation if 2,3 and5 always occur in each number
Pujayiya , 8 Years ago
Grade 10
10 grade maths> How many number of five digits can be for...
1 Answers
Arun
Last Activity: 8 Years ago
Let's think of 2 cases.
1. A formed number
has 2, 3 or 5 in 10 thousands' place-
Out of remaining 4 places, 2 places are fixed. Now, out of remaining 2 places, one can be filled with 7 digits ( except 2, 3 & 5) & other with 6 digits ( One which fills first unknown place is excluded.).
Position of these 4 digits can be reversed in 4! ways.
But doing this, same condition occurs twice. E.g. We have selected (6,7) & (7,6) as well.
So, we have to divide the total number by 2.
As 5th place can be filled with 2, 3 or 5, we have
Numbers formed by condition 1 = 4!*7*6*3/2=63*4!
2. A formed number has no 2, 3 or 5 in its 5th place-
5th place can be filled by 6 digits
(excluding 2,3,5 & 0). If 0 is in 5th place, it will be a 4 digited number. Remaining unknown place can be filled by 6 digits (excluding 2,3,5 & one which is in 5th place).
If we keep 5th place fixed, other 4 digits can be reversed in 4! ways.
Numbers formed by condition 2 = 4!*6*6
Total numbers = 4!*63+4!*36
=24×99=2376.
1. A formed number
has 2, 3 or 5 in 10 thousands' place-
Out of remaining 4 places, 2 places are fixed. Now, out of remaining 2 places, one can be filled with 7 digits ( except 2, 3 & 5) & other with 6 digits ( One which fills first unknown place is excluded.).
Position of these 4 digits can be reversed in 4! ways.
But doing this, same condition occurs twice. E.g. We have selected (6,7) & (7,6) as well.
So, we have to divide the total number by 2.
As 5th place can be filled with 2, 3 or 5, we have
Numbers formed by condition 1 = 4!*7*6*3/2=63*4!
2. A formed number has no 2, 3 or 5 in its 5th place-
5th place can be filled by 6 digits
(excluding 2,3,5 & 0). If 0 is in 5th place, it will be a 4 digited number. Remaining unknown place can be filled by 6 digits (excluding 2,3,5 & one which is in 5th place).
If we keep 5th place fixed, other 4 digits can be reversed in 4! ways.
Numbers formed by condition 2 = 4!*6*6
Total numbers = 4!*63+4!*36
=24×99=2376.
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